Measuring the quality of real approximation

Question

Let $r\in [0,1]\setminus\mathbb{Q}$, let $\mathbb{N}$ denote the set of non-negative integers and let $\mathbb{N}_+=\mathbb{N}\setminus\{0\}$. For $n\in\mathbb{N}_+$ let $$\alpha_r(n)=\min\{\big|r-\frac{m}{n}\big|: m\in\mathbb{N}\}$$ be the best approximation of $r$ that can be obtained using $n$ as the denominator.

For $k\in\mathbb{N}_+$ we say that $r\in[0,1]\setminus\mathbb{Q}$ is $k$-good if there is $c\in\mathbb{N}_+$ as well as a "starting index" $N_0\in \mathbb{N}_+$ such that $$\alpha_r(n) < c \cdot \frac{1}{n^k} \;\text{ for all } n\geq N_0.$$

Note that for all irrational $r$ in the unit interval, we have $\alpha_r(n) < \frac{1}{n}$ for all $n\in\mathbb{N}_+$, so every such $r$ is $1$-good.

For which integers $k \geq 2$ is there a $k$-good irrational number $r\in[0,1]\setminus\mathbb{Q}$? Is there even such an $r$ that is simultaneously $k$-good for all integers $k\geq 2$? (Only the first question needs to be answered for acceptance.)

Answer 1

This certainly can't happen for any $k > 1$. (There is no reason to assume that $k$ is an integer rather than real number). In fact, it can't even happen if you just restrict the $n$ to sufficiently large powers of two.

Suppose you know that for every sufficiently large integer $i$ there is an integer $m$ such that

$$\left|r - \frac{m_i}{2^i}\right| < \frac{c}{2^{ik}}.$$

Compare this appoximation with the approximation for $n = 2^{i+1}$. We have

$$\left|r - \frac{m_{i+1}}{2^{i+1}}\right| < \frac{c}{2^{(i+1)k}}$$

By the triangle inequality

$$\left| \frac{m_i}{2^i} - \frac{m_{i+1}}{2^{i+1}} \right| < \frac{c}{2^{ik}} + \frac{c}{2^{(i+1)k}}.$$

If $k > 1$, then for large enough $i$ (depending only on $c$) the RHS will be strictly less than $1/2^{i+1}$. On the other hand the left hand side is equal to

$$\left| \frac{2m_{i} - m_{i+1}}{2^{i+1}} \right|$$

and this is at least $1/2^{i+1}$ if it is non-zero. Hence the only way the inequalities can hold for all sufficiently large $i$ is if eventually we have $2 m_i = m_{i+1}$ for all $i \ge i_0$. But then our sequence of rational approximations to $r$ are all equal to the (rational) number $m_{i_0}/2^{i_0}$, which implies that $r$ is rational.